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Given the Sturm-Liouville problem on $a < x < b$: $$y" + \lambda y = 0$$ satisfying boundary conditions: $$cos(\alpha)y(a) - Lsin(\alpha)y'(a) = 0$$ $$cos(\beta)y(b) - Lsin(\beta)y'(b) = 0$$

I'm trying to prove that if you have two eigenfunctions corresponding to the same eigenvalue, they will be constant multiples of each other. The book I'm reading accomplishes this proof by letting $g(x), h(x)$ be solutions to the SL problem. For $\alpha = 0$, $g(a) = h(a) = 0$. Define: $$f(x) = g'(a)h(x) - h'(a)g(x)$$ $f(x)$ also satisfies the equation and initial conditions $f(a) = f'(a) = 0$. The part I'm having trouble with is that they then conclude that $f(x) = 0$. I don't understand how this is immediately obvious, and my book gives no explanation for that fact.

I can only imagine that this conclusion is a result of the uniqueness theorem. Since $f(x) = 0$ solves equation, by uniqueness theorem, it must be the only solution for these initial conditions. If that is true, I undrstand how the rest follows. But I am troubled by the lack of explanation, as if there is something else demanding that $f(x) = 0$

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The general solution of $y" + \lambda y = 0$ is given by $$ f(x)=A e^{i\sqrt\lambda x}+B e^{-i\sqrt\lambda x}$$

so $$ f'(x)=i\sqrt\lambda \left[A e^{i\sqrt\lambda x}-B e^{-i\sqrt\lambda x} \right]$$

The conditions $f(0)=f'(0)=0$ reduce to ...

$$\begin{eqnarray*} A+B=&0 \\i\sqrt\lambda \left(A -B \right) =& 0 \end{eqnarray*}$$

the only solution to this system is $A=B=0$ hence $f(x)=0$